Ro Linked List Python

linked_list/linked_list.py

@@ -1,79 +1,118 @@
 
-# Defines a node in the singly linked list
+
 class Node:
 
     def __init__(self, value, next_node = None):
         self.value = value
         self.next = next_node
 
-# Defines the singly linked list
 class LinkedList:
     def __init__(self):
-      self.head = None # keep the head private. Not accessible outside this class
-
-    # returns the value in the first node
-    # returns None if the list is empty
-    # Time Complexity: ?
-    # Space Complexity: ?
+      self.head = None 
     def get_first(self):
-        pass
-
+        if self.head:
+            return self.head.value
+        return None
 
-    # method to add a new node with the specific data value in the linked list
-    # insert the new node at the beginning of the linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(1)
+    # Space Complexity: O(1)
     def add_first(self, value):
-        pass
+        self.head = Node(value, self.head)
 
-    # method to find if the linked list contains a node with specified value
-    # returns true if found, false otherwise
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(1)
+    # Space Complexity: O(1)
     def search(self, value):
-        pass
+        current = self.head
+
+        while current:
+            if current.value == value:
+                return True
+            current = current.next
+
+        return False
 
-    # method that returns the length of the singly linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def length(self):
-        pass
+        count = 0
+        current = self.head
+        while current:
+            count += 1
+            current = current.next
+
+        return count
 
-    # method that returns the value at a given index in the linked list
-    # index count starts at 0
-    # returns None if there are fewer nodes in the linked list than the index value
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def get_at_index(self, index):
-        pass
+        current_index = 0
+        current = self.head
+        while current_index < index and current:
+            current_index += 1
+            current = current.next
+
+        if current_index == index and current:
+            return current.value
+        return None
 
-    # method that returns the value of the last node in the linked list
-    # returns None if the linked list is empty
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def get_last(self):
-        pass
+        if not self.head:
+            return None
+
+        current = self.head
+        while current.next:
+            current = current.next
 
-    # method that inserts a given value as a new last node in the linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+        return current.value
+
+    # Time Complexity: O(n); more than 1 approach; same complexity 
+    # Space Complexity: O(n); more than 1 approach; same complexity 
     def add_last(self, value):
-        pass
+        if not self.head:
+            self.add_first(value)
+        else:
+            new_node = Node(value)
+            current = self.head
+            while current.next:
+                current = current.next
+            current.next = new_node
+
 
-    # method to return the max value in the linked list
-    # returns the data value and not the node
     def find_max(self):
-        pass
+        if self.head == None:
+            return None
+
+        current = self.head
+        max = self.head.value
+        while current:
+            if current.value > max:
+                max = current.value
+            current = current.next
+
+        return max
 
-    # method to delete the first node found with specified value
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def delete(self, value):
-        pass
+        if not self.head:
+            return
+
+        if value == 0:
+            self.head = self.head.next
+
+        current = self.head
+        current_index = 0
+        while current.next and current_index < value - 1:
+            current = current.next
+            current_index += 1
+
+        if current.next:
+            current.next = current.next.next
 
-    # method to print all the values in the linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(1) after element is found; O(n) prior
+    # Space Complexity: O(1)
     def visit(self):
         helper_list = []
         current = self.head
@@ -84,45 +123,53 @@ def visit(self):
 
         print(", ".join(helper_list))
 
-    # method to reverse the singly linked list
-    # note: the nodes should be moved and not just the values in the nodes
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def reverse(self):
-        pass
-
-    ## Advanced/ Exercises
-    # returns the value at the middle element in the singly linked list
-    # Time Complexity: ?
-    # Space Complexity: ?
+        if not self.head:
+            return
+
+        current = self.head
+        previous = None
+
+        while current:
+            next = current.next
+            current.next = previous
+            previous = current
+            current = next
+
+        self.head = previous
+
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def find_middle_value(self):
-        pass
+        if not self.head:
+            return None
+
+        return self.get_at_index(int(self.length() / 2))
 
-    # find the nth node from the end and return its value
-    # assume indexing starts at 0 while counting to n
-    # Time Complexity: ?
-    # Space Complexity: ?
+    # Time Complexity: O(n)
+    # Space Complexity: O(n)
     def find_nth_from_end(self, n):
-        pass
-
-    # checks if the linked list has a cycle. A cycle exists if any node in the
-    # linked list links to a node already visited.
-    # returns true if a cycle is found, false otherwise.
-    # Time Complexity: ?
-    # Space Complexity: ?
-    def has_cycle(self):
-        pass
-
-    # Helper method for tests
-    # Creates a cycle in the linked list for testing purposes
-    # Assumes the linked list has at least one node
-    def create_cycle(self):
-        if self.head == None:
-            return
+        if not self.head:
+            return None
 
-        # navigate to last node
         current = self.head
-        while current.next != None:
+        trail = None
+        spacing = 0
+        while current and spacing < n:
             current = current.next
+            spacing += 1
+
+        if not current:
+            return None
+
+        trail = self.head
+        while current.next:
+            current = current.next
+            trail = trail.next
+
+        return trail.value
 
-        current.next = self.head # make the last node link to first node
+    # Time Complexity: O(log n)
+    # Space Complexity: O(log n)